# Every Closed Set In $R^1$ is intersection of countable collection of open set.

This is question I tried to solve as follows
Consider A be closed set in R Therefore R\A is Open set .Now By Representation theorem of open set ,Every open set in R can be written as union of countable collection of disjoint open interval.
R\A=$\cup I_n$ Where $I_n$ is open interval where n is form countable index set.
R(R\A)=A=R\ $\cup I_n$ =$\cap R\I_n$ which implies A is countable intersection of Closed set .
I had to prove that it is intersection of countable intersection of open set but I got other answer Where is my mistake in argument ? Any Help will be appreciated

• The mistake in your argument is... that you didn't prove what you set out to prove. That a closed set is the intersection of a countable collection of closed sets is clear: namely, it is the intersection of countably many copies of itself. – Mees de Vries May 3 '18 at 12:22
• Hint for solving this: can you do it for a closed interval? Can you do it for two disjoint closed intervals? Can you generalize? – Mees de Vries May 3 '18 at 12:22
• As Closed interval in R are compact so I can have its finite sub cover of any open cover which can cover that .This I can extend to to finite disjoint closed interval .But Question is about intersection ,That I am not Getting. – SRJ May 3 '18 at 12:26

This works in every metric space $(X, d)$:

Hints: Let $A$ be closed in $X$, then

1. $A=\{x:d(x, A) =0\}$
2. Consider $A_n:=\{x:d(x, A) <\frac1n\}$.

where $d(x, A)$ denotes the distance of point $x$ to set $A$, i.e. $$d(x, A) =\inf_{a\in A} d(x, a)$$ And specifically for $\Bbb R$, the distance function is given by $d(x, y) :=\vert y-x\vert$.

For $k=1,2,...$, cover the real line with the intervals $(n/2^k-1/2^{k+1},(n+1)/2^k+1/2^{k+1})$, for $n\in\mathbb{Z}$. Define $U_k$ to be the union of those intervals in this collection that intersect your closed set $C$.

The claim is that $\cap_k U_k=C$. Clearly $C\subset\cap_k U_k$ because $C\subset U_k$ for all $k$.

Assume that $x\in\cap_k U_k$. Then for each $k$ there is an $n_k\in\mathbb{Z}$ such that $x\in(n_k/2^k-1/2^{k+1},(n_k+1)/2^k+1/2^{k+1})\subset U_k$. But this interval contains some point $c_k\in C$ by definition of $U_k$. Therefore, $c_k\to x$ because the sizes of those intervals $1/2^k-1/2^{k+1}\to0$. Therefore $x\in C$ because $C$ is closed.